User:Masoud Sheykhi

'''Copper Smelting& Converting

Sar Cheshmeh copper complex Copper converting

"A calculation on copper converters"                              '''

Masoud Sheykhi

Process inspection

Smelter Process

Converter inspection

Technical_inspection@nicico.com

Introduction

Copper ores usually smelted to matte, which for purposes of

calculation is usually taken as a simple molten solution of

CU2S and FeS. The percentage of copper in the matte

commonly termed the "grade" of the matte, thus determines

the percentage composition in CU ,Fe, and S of this ideal

matte ; the percentage of CU multiplied by 160/128 give the

percentage of CU2S, and the balance is FeS.High- grade

mattes commonly have part of their copper content present as

free CU. The action in a copper smelting furnace and also

in a converter is governed by the fact that sulphur has a

greater affinity for copper than for iron. Though the heat

of formation of FeS is greater than that of CU2S, the

oxidation of FeS has greater free energy change than the

oxidation of CU2S ,and the reaction 2CUO + FeS = CU2S+FeO

will take place from left to right as shown. The operation

of copper converter is divided into two stages. the first ,

or slag- forming ,stage consists in the oxidation of FeS to

FeO with SiO2 added as flux:

2FeS +3O2 = 2FeO +2SO2

x FeO+ySiO2 = xFeO.ySiO2

At the end of the first stage the slag is poured off;the

material remaining consists mostly(theoretically entirely)of

CU2S ,Called"white metal". The second stage is the oxidation

of CU2S to Blister copper:

CU2S + O2 = 2CU +SO2.

Theoritically no slag is made during the seconde stage, and

this may be assumed to be the case in calculations.

Calculations

In the room temperature we have:

CU2S + O2 = 2CU +SO2

In 1100[0c] we have;

Heat contents of reactants ;

in 160kg CU2S = 160*0.131*1100

=                              23050 Kcal

Laten heat of smelting       = 160* 34.5

=                          5520 Kcal

in 22.4 M3 O2 = 22.4*(0.302*0.000022*1100)*1100 = 8030 Kcal Total heat of reactants =                                                     36600Kcal

Heat content of products;

in 128 Kg CU = 128*(0.0916+0.0000125*1083)*1083 =                14600 Kcal in 128 Kg CU = 128*41.8 =                                                                  5360 Kcal in (1083[0c]-1100[0c])        =                                                                 180   Kcal in 22.4 M3 SO2       = 22.4*(0.406+0.00009*1100)*1100 =              12440  Kcal -

Total heat in products

=                                                                        32580Kcal

Reaction heat in 1100[0c] = -51990-36600+32580 =                        -56010Kcal

Heat in nitrogen = (79/21)*22.4*(0.302+0.000022*1100)*1100

= 30250Kcal

net heat released in converter =  56010-8030-30250

=                    17730 Kcal

References

[1] A.B.,"Metallurgical problems"McGRAW-HILL,INC,1943.

-

Sar Cheshmeh copper converters

"Calculations on Scrap addition during copper blowing "

Masoud Sheykhi'''

Process inspection

Smelter Process

'Converter inspection'

Technical_inspection@nicico.com

Calculations

q[Watts/cm2] |(copper blowing)[1] = 5.76*0.0.357*[(T2/1000)4-

(T1/1000)4]=

2.06*[(T2/1000)4-(T1/1000)4]

For copper blowing we have:

Qc[Mcal/cm2.sec] = 0.012*0.5*(t2-t1)^1.233

if T1 = 1373[0K], T2 = 298[0K] ,S = 90000 cm2 , then ;

q =9.42Mcal/min

if T1 = 1443[0K], T2 = 298[0K] ,S = 90000 cm2 , then ;

q =11.5 Mcal/min if t1 = 25[0c], t2 = 200[0c] ,S = 1380000cm2 ,then;

Qc  = 3.498 Mcal/min

q+Qc = 9.42+3.498 = 12.92 Mcal/min

Daily[2]; 1110 Tons of matte with 41.4%(take 50% for

calculations) CU charged in 3 converters, copper blowing

times = 113 minutes ,each converter cycle times = 9 hr.

Hence; each converter has 2.66 cycles per day and copper

blowing times for 3 converter is 15.029hr/day. thus;

released heat during copper blowing[3] = (17.73Mcal/160Kg

CU2S)*[(1110000/2)Kg CU2S/day]*[ day/15.029hr]*[hr/60min] =

68.2 Mcal/min.

heat accumulated in each converter =( 68.2/3)-12.92=

9.813Mcal/min.

heat accumulated in 3 converters = 3*9.81 = 29.44 Mcal/min.

heat carried by blister copper for changing its temperature

from 25[0c] to 1100[0c] = (14.6+5.360+0.18)/128 = 0.157

Mcal/Kg.

Daily production[2] ; are 420Tons blister copper.

(0.151Mcal/Kg)*[420000/(3*15.029*60)] =

24.428Mcal/min.converter

24.428/accumulated heat per converter = 24.428/9.813 = 2.5

Hence ; Max of Scrap% = 9.813*100/(2.5*24.42) = 16

Now ,we check the above calculations ;

We have[3];

32.580-12.44-[3.498*17.73/68.2] = 19.23Mcal/160Kg CU2S =

24.66Mcal/min.converter that near to 24.428 .hence

calculations are true.

Min of scrap% =[9.813/(24.66*2.50)]*100 =15.92

Now,We take the average of 24.66 and 24.428 i.e;24.54

24.54/9.813 = 2.50

Hence ;

Average of Scrap% =[9.813/(24.54*2.50)]*100 = 15.995.

References

[1]M.Sheykhi "Emissivity calculation&convection heat

transfer coefficient on sarcheshmeh copper complex's

converters", technical inspection division, sarcheshmeh

copper complex ,kerman,iran ,2003.

[2]Sarcheshmeh "Smelter operating manual ,section 7 ;

Converting" ,Paeson Jurdan Int .corp,1977.

[3] M.Sheykhi " A calculation on  copper  converters",

technical inspection division, sarcheshmeh copper

complex ,kerman,iran ,2004.

-

'''Emissivity calculation & convection heat transfer

coefficient on Sar Cheshmeh copper complex's Converters

Masoud Sheykhi

Process inspection

Smelting Process

Converter inspection

Technical_inspection@nicico.com

Calculations

Converter heat losses equation[1] ;

Q[kcal/m2h]= [4.88*e*(T/100)^4]*S ,Let; e = .80

Let assume that converter has 4 meter diameter and 9 meter

length then;

Q[kcal/min]= [0.583*(T/100)^4]

Let temperature of the melt surface be 1200 0c then;

q[Mcal/min] = [4.88*0.80*{1200+273)/1000}^4]*9/(60*1000)=

27.568

Let temperature of the shell surface be 200 0c then;

q'[Mcal/min] = [4.88*0.80*{200+273)/1000}^4]*138/(60*1000)=

4.498

Q' = q'+ q = 27.568+4.498 = 32.066

(q/Q' )*100 = (27.568/32.066)*100=85.97

(q'/Q')*100 = (4.498/32.066)*100=14.03

Qc = convection heat losses from converter shell = 2.66

Mcal/min[2].Thus;

4.498-2.66 = 1.83 Mcal/min is the radiation heat losses from

converter shell.

we have [2];

Qc[Mcal/cm2.sec] = n*(t2-t1)^1.233

t1[0c] ; temperature of air sourrounding converter shell.

t2[0c] ; temperature of converter shell

n ; is a constant depending on geometry and condition of surface.

A = 1380000cm2

Hence;

n = 0.012 Mcal.cm-2.min-1.[c0]-1.233

Assume that total radiation heat losses from the converter mouth, thus;

q[Mcal/min] = 27.568+1.83 = 29.398.

we have[3] ;

q[Watts/cm2] = 5.76*e'*[(T2/1000)4-(T1/1000)4]

T2=1200+273=1473 [0K] T1=25+273=298[0K]

29.398[Mcal/min] = 22.78[Watts/cm2]

Hence, we correct;

e=e' = 0.84

Radiation heat losses during slag blowing[Mcal/min] = 5.416

Radiation heat losses during copper blowing[Mcal/min] = 4

Convection heat losses during converter blowing[Mcal/min] =

2.66

Mean radiation heat losses during converter blowing

[Mcal/min] = 4.708>4.498

Sarcheshmeh converter Slag blowing times[hr][4] = 4.52

Sarcheshmeh converter Copper blowing times[hr][4] = 1.88

Sarcheshmeh converter Total blowing times[hr] = 6.4*[5.416/(5.416+4)]*100 = 57.52

100-57.52 =42.48

Therefore, during copper blowing we have;

e = 0.4248*0.84 =0.3568

and during slag blowing ;

0.5752*0.84 = 0.4832

Hence;

q[Watts/cm2]|(slag blowing) = 5.76*0.4832*[(T2/1000)4-

(T1/1000)4]= 2.78*[(T2/1000)4-(T1/1000)4]

q[Watts/cm2] |(copper blowing) = 5.76*0.0.3568*[(T2/1000)4-

(T1/1000)4]=

2.06*[(T2/1000)4-(T1/1000)4]

[2.66/(2.66+2.66)]*100 = 50

100-50 = 50

Hence;

Qc[Mcal/cm2.sec] = 0.12*0.50*(t2-t1)^1.233 during slag blowing

Qc[Mcal/cm2.sec] = 0.12*0.50*(t2-t1)^1.233 during copper blowing

References

[1] S.GOTO ; The Application of thermodynamic calculations

to converter practice ,paper presented at AIME annual

meting ,1979.

[2] Nickel Converter " Material And Heat Balance

report" ,Outokumpu Oy co ,1968.

[3] Ann .chim .phys.,7(1817).

[4] Sarcheshmeh Smelter operating manual ,section 7; Converting ,Parson Jurdan Int .corp ,1977.

- Sar Cheshmeh copper converters

'''" Calculations on Cold material addition during Slag

blowing "

Masoud Sheykhi

Process inspection

Smelter Process

Converter inspection

Technical_inspection@nicico.com

Calculations

q[Watts/cm2]|(slag blowing)[1] = 5.76*0.4832*[(T2/1000)4-

(T1/1000)4]=

2.78*[(T2/1000)4-(T1/1000)4]

Qc[Mcal/cm2.sec]|(slag blowing) = 0.012*0.50*(t2-t1)^1.233

if T1 = 1477[0K], T2 = 298[0K] ,S = 90000 cm2 , then ;

q =17.185Mcal/min.converter.

if t1 = 25[0c], t2 = 200[0c] ,S = 1380000cm2 ,then; Qc =

3.498 Mcal/min.converter.

q+Qc = 17.185 + 3.498 = 20.683 Mcal/min.converter = 60.6

Mcal/min.3 converter.

Reverts used during slag blowing have the following analyses

[2];

Element/compound : %

CU : 25

Fe :22                                                                                                             S :8

Sio2:27

Cao:3

Al2o3:3

Others:12 -

Terminology for heat characteristics of components of the

reverts(Cold materials),

Say ;[3];

Tm[0c] ; melting point

Lf[Kcal/Kg]= heat of fusion

Cm[Kcal/Kg]; specific heat

QLMP[Kcal/Kg]; Total liquid heat content at melting point

Tb[0c] ; boiling point

Lg[Kcal/Kg] ; evaporation laten heat

-

CU :Tm=1084 ;Lf=57.6 ,Cm=0.092 +0.0000125t ,QLMP = 172

,Cm(liquid) = 0.112

Fe:

0-600[0c] ; Tm = 0.104+0.000064t ,QLMP = 85*

0-760[0c] ; Tm = 0.180+0.0003(t-6) ,QLMP = 121*

760-910[0c]; Tm = 0.320-0.000053(t-760), QLMP = 161*

910-1400[0c] ; Tm = 0.160+0.00001(t-910) ,QLMP = 244*

1400-1534[0c] ;Cm = 0.185 ,QLMP = 334*

S :

Tm = 119  ,Lf = 9.2   ,Tb =445  ,Lg = 79

Sio2 :

Quartz ; Tm = 1470, Lf = 57

Cristobalite; Tm = 1700, Lf = 35

Cao :

Tm = 1700, Lf = 35

Al2O3 :

Tm = 2045, Lf = 250

-


 * heat content of solid iron above [0c].

heat content of reverts(Cold materials) =

0.27(0.092+0.0000125*1084)+0.27*57.6+0.27*0.112(1100-1084)

+0.24(85+121+161+(0.160+0.00001(1100-910)*1100]+0.10

[30.60+193+0.153(1100-445)]

+0.29*0.19*1100+0.05*0.24*1100+.05*0.28*1100] =

299.75396 Kcal/Kg Cold materials, from which; 59.5 Kcal/

Kg Cold materials belongs to copper in the reverts(Cold

materials). Hence; 59.5/299.75396 =

0.198 =mass of copper / mass of cold materials.

2Fes + 3O2 = 2Feo + 2So2 ,Reaction heat at 298 degree kelvin

= -230000

Kcal/2Kgmol Fes.

2Feo + Sio2 = 2Feo.Sio2, Reaction heat at 298 degree

kelvin = -10000

Kcal/Kgmol sio2.

2Fes + 3O2 +Sio2 = 2Feo.Sio2 +2So2, Reaction heat at 298

degree kelvin

= -240000

Kcal/Kgmol sio2.

heat content at 1200[0c];

in 176 Kg Fes = 176(0.1355+0.000078*1200)1200= 48.4Mcal/176

Kg Fes

melting leaten heat = 176*57

=                                 10Mcal/176Kg Fes

in 60Kg Sio2 = 60*0.264*1200 =                                      19Mcal/176Kg Fes

in 3*22.4M3 O2 = 3*22.4(0.302+0.000022*1200)1200 =26.4

Total heat of reactants =                                                   104

Mcal/176Kg Fes.

in 2*22.4 M3 So2 = 2*22.4(0.406+0.00009*1200)1200=27.6

Mcal/176 Kg Fes.

in 264 Kg(2Feo.Sio2) = 264*0.227*1200 =

71.9 Mcal/176Kg Fes.

Total heat of products

=                                             99.5

Mcal/176Kg Fes .

Reaction heat at 1200[0c] = -240-104+99.5 = -244.4 Mcal/176

Kg Fes.

Heat in nitrogen = (79/21)*3*22.4(0.302+0.000022*1200)1200 =

99.6 Mcal/176Kg Fes.

Net heat released = 244.4-10-99.6 = 134.7 Mcal/176 Kg Fes.

Slag blowing times in each sarcheshmeh copper converter[2] =

271minutes.

Daily we have 2.66 cycles for each converter[2].

Hence; daily slag blowing times for 3 converter are 36 hr.

134.7[Mcal/176 Kg Fes]*1110000[Kg matte/day]*0.50[Kg

Fes/matte]/(36[hr]*60[min]) = 196.66 Mcal/min.

heat in nitrogen[Mcal/min] = 99.6*196.66/134.7 = 145.4.

Accumulated heat in converters during slag blowing = 196.66-

145.4-60.6 = 9.34 Mcal/min. on the other hand;

Accumulated heat in converters during slag blowing[Mcal/min]

=

m(cold materials)[Kg/min] * 0.29975396 Mcal/Kg thus;

m(cold materials) = 31.13 Kg/min.

where m(cold materials) be the mass of cold materials needed

for slag blowing.

Cold materials needed for one day = 31.13*36*60/1000 =

67.248 Tons/day.

Scrap % [4] = 16

Blister production[2] = 420 Tons /day

Scrap needed = 420*0.16 = 67.2 Tons/day

67.2 * mass of cold materials/mass of copper = 67.2 *0.198 =

13.3056 Tons /day cold materials.

thus; Total reverts (Scrap + cold materials ) needed for Slag and

copper blowing =

13.3056 +67.248 =80.6 Tons /Day.

By [2] we need 80 Tons ;reverts /day for Slag and copper

blowing .therefor calculations agreed with designed criteria. References

[1]M.Sheykhi "Emissivity calculation&convection heat

transfer coefficient on     sarcheshmeh copper complex's

converters", technical inspection division, sarcheshmeh

copper complex ,kerman,iran ,2003.

[2]Sarcheshmeh "Smelter operating manual ,section 7 ;

Converting" ,Parson Jurdan Int .corp,1977.

[3] Liddell ,Donald M ."The metallurgists, and chemists

Hand Book" ,3d ed.,McGraw-Hill Book ,Inc., New York ,1930.

[4] M.Sheykhi " Sarcheshmeh copper converters- Calculations

on Scrap addition during copper blowing ", technical

inspection division, sarcheshmeh copper

complex ,kerman,iran ,2004.

Common cause of Regrinding cyclone pumps shaft failure  Sar Cheshmeh Copper Complex

Masoud Sheykhi Common cause of pump shaft failure is due to breakage caused by the pump ingesting hard particles or foreign objects(Balls,etc).

If the pump shaft does not break from the sudden impact, it could still fail in the near future due to stress cracks. In some instances the hard particles(Balls,etc) will distort the impeller and cause the pump shaft to have excessive run out.

This will result in the impeller rubbing and literally gouging or imbedding itself into the pumps wear ring liner. This excessive force can easily break the pump shaft.

While it is not possible to eliminate this problem, regular inspection and maintenance will extend life and allow for replacement before it becomes a problem and ruins & outing.